How tio use like operator in textitem to get the name
Oracle forms6i
Hai All
I have created an form to create an hierarchy list. I that I had an Text_item how can i use like operator to get the name to select. which trigger i need to use. when i select the name i need to get his employee code and other details
Thanks In Advance
Srikkanth.M
A is the head of Dept 10 and three person are under A namelyHow will you know that B, C and D are under A is there any parent child relationship between Leader and Employee code? If yes, Then you can create the relationship between blocks no need to set where_clause programatically. So, relationship can be like...
leader_block.emp_code = emp_block.leader_code
AND leader_blokc.dep_code = emp_block.dep_codeSo, when you will query data in LEADER block it will show employees which are under LEADER which you queried in LEADER block.
Or if there is no relationship then how will you know that which employee under which leader?
b 002 10 1
c 003 10 1
d 004 10 1
When i enter the name of head corresponding other members in the department will need to display in the
Tabular format Or if you want to query the records only for department relation. Then as you said you create two blocks then create create the relationship using dep_code between those blocks.
like...
leader.dep_code = emp_block.dep_code-Ammad
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sql syntax
SQL>SELECT customer_name FROM customer_header
WHERE customer_name LIKE 'B%' ; customer_name
Bala murali
Babu
Basker
plsql syntax
PROCEDURE pro_custheader_like ( v_cname IN varchar2
,answer OUT type_refcur_customer) IS
BEGIN
OPEN answer FOR
SELECT customer_name FROM customer_header
WHERE customer_name LIKE ( ' v_cname ' );
END pro_custheader_like;
execution command
sql>variable answer refcursor;
sql>set serveroutput on
sql>exec package_name.pro_custheader_like( 'R',:answer);
plsql successfully completed
sql>print :answer
no row selected
by
balamuralikrishnan.splsql syntax
PROCEDURE pro_custheader_like ( v_cname IN
varchar2
,answer OUT
type_refcur_customer) IS
N
OPEN answer FOR
SELECT customer_name FROM customer_header
WHERE customer_name LIKE ( v_cname );
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Try it without any quotes. And, let us know your feedback.
Regards.
Satyaki De.
Message was edited by:
Satyaki_De -
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Hi I am trying to use like operation in loop condition. the code is as follows:
loop at lt_mara into l_mara where l_mara like '%XXX'.
endloop
when i have the like operator like above its giving me an error that i cannot use like operator. Is there anyway that i can use like operation in loop at condition,
Thanks in advanceusing the ranges with CP option we can do that..
Populate the ranges mentioned below..
data: r_matnr type range of matnr,
w_matnr like line of r_matnr.
w_matnr-low = '%XXX'.
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I want to find the value of ' % ASSOC%' in the name and replace it to 'Assoc.' using ifthenelse.
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if you have data ASSOC1,ASSOC2
and you want Output like Assoc.1,Assoc.2.
In this condition it is not possible with ifthenelse function
it will return output as Assoc.
Example Source:
If we use if then else condition. Then it will replace all the Data to Assoc.
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Hi, I need to execute a query with the LIKE operator, but using a PreparedStatement. Can I do this, and if so what must my SQL look like with the wildcard characters '%' or '_'?
normal PS example: conn.prepareStatement("select * from mytable where name like ?");
If I try: conn.prepareStatement("select * from mytable where name like ?%");
I get: ORA-00911: invalid character
If I try: conn.prepareStatement("select * from mytable where name like '?%'");
I get: ORA-01006: bind variable does not exist
I must use a PreparedStatement, as my variable may contain illegal characters (like '), and using PreparedStatement.setString(1, var) will automatically escape it for me.
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when my variable contains a '%' symbol, the PreparedStatement will NOT escape it, and thus it is treated like a wildcard. Therefore, I can just do
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HI all,
I have a requirement as follows
EMPID ENAME JOB SAL
10 RAJ KAMAL MANAGER 25000
20 KAMAL RAJ NAYAK CLERK 4000
30 NARENDAR GUPTA ANALYST 20000
40 ASHUTOSH GUPTA DEVELOPER 10000
50 ASHUTOSH NAYAR PROGRAMMER 15000
i am searching enames such that i need to get the whole name even if i search with just a single LETTER/WORD immaterial of the order they exist in the name for which i need to get the whole name.(INFACT WORD comparision)
ex:
1) select * from emp where ename like '%ka%'
i am getting:
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20 KAMAL RAJ NAYAK CLERK 4000
This is not what i need i need just word camparision not letters comparision let me tell you...
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Even for this query ill get the same output.. this is not my option to go..
I need just word comparision with starting letter and immaterial of the word position in the name
If it is possible to do with query please let me know..
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SQL> ed
Wrote file afiedt.buf
1 with e as (select 10 as empid, 'RAJ KAMAL' as ename, 'MANAGER' as job, 25000 as sal from dual union all
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3 select 30, 'NARENDAR GUPTA', 'ANALYST', 20000 from dual union all
4 select 40, 'ASHUTOSH GUPTA', 'DEVELOPER', 10000 from dual union all
5 select 50, 'ASHUTOSH NAYAR', 'PROGRAMMER', 15000 from dual)
6 -- END OF TEST DATA
7 select *
8 from e
9* where regexp_like(ename,'(^| )KA( |$)')
SQL> /
no rows selected
SQL> ed
Wrote file afiedt.buf
1 with e as (select 10 as empid, 'RAJ KAMAL' as ename, 'MANAGER' as job, 25000 as sal from dual union all
2 select 20, 'KAMAL RAJ NAYAK', 'CLERK', 4000 from dual union all
3 select 30, 'NARENDAR GUPTA', 'ANALYST', 20000 from dual union all
4 select 40, 'ASHUTOSH GUPTA', 'DEVELOPER', 10000 from dual union all
5 select 50, 'ASHUTOSH NAYAR', 'PROGRAMMER', 15000 from dual)
6 -- END OF TEST DATA
7 select *
8 from e
9* where regexp_like(ename,'(^| )KAMAL( |$)')
SQL> /
EMPID ENAME JOB SAL
10 RAJ KAMAL MANAGER 25000
20 KAMAL RAJ NAYAK CLERK 4000
SQL> -
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reagrdskindly press Shift+F1 at a time you face this error to see the exact Oracle error message.
and provide us with that detail
and its better if you start new topic for that error... because that will be new error,,,
-- Aamir Arif
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Here you go. In 'SQL Editor' in PL/SQL Developer:
Pulled this off the PL/SQL Developer Manual:
Escape character
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LINE 1 : String temp = "AA";
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